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Singular extremal of a constrained variational problem

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Consider the following constrained variational problem: $$\min_{u \in H^{1}(I)} \{\mathcal{F}(u) : u(\pm 1) = 1, \mathcal{G}(u) = 1/3 \},$$ where $I = [-1, 1] \subseteq \mathbb{R}, H^1 (I) := H^{1, 2}(I)$ is a one-dimensional Sobolev space and $$\mathcal{F}(u) := \int_I |u'(x)|^2 dx, \;\;\;\;\mathcal{G}(u) := \int_I \left(x^2 u(x) -\frac{u(x)^3}{3} \right)dx$$ are respectively the objective and constraint functionals.

If $u$ is a $\mathcal{G}$-regular extremal, then the Euler-Lagrange equation leads to the following boundary value problem: $$u''(x) = \lambda(u^2(x)-x^2), \;\;\; u(\pm 1) = 1.$$However, it turns out that there is also a $\mathcal{G}$-singular extremal (for which $\frac{d}{d\epsilon} \mathcal{G}(u + \epsilon \psi)|_{\epsilon = 0} = 0 \; \forall \psi \in C_c^{\infty}(\mathring{I}))$, namely $u(x) = |x|$. The question is the following: is $u(x) = |x|$ a minimum of the described problem?

The approaches I have tried involve trying to study the previous differential equation or searching for ways to modify the singular solution to get lower values for the energy, but neither way has yielded results.


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